List Grades (25)

标签：技术   syn   col   return   style   you   with   分享图片   amp   

题目描述

Given a list of N student records with name, ID and grade. 
You are supposed to sort the records with respect to the grade in non-increasing order, 
and output those student records of which the grades are in a given interval.

输入描述:

Each input file contains one test case.  Each case is given in the following format:
N
name[1] ID[1] grade[1]
name[2] ID[2] grade[2]
... ...
name[N] ID[N] grade[N]
grade1 grade2
where name[i] and ID[i] are strings of no more than 10 characters with no space, grade[i] is an integer in [0, 100], 
grade1 and grade2 are the boundaries of the grade‘s interval.  It is guaranteed that all the grades are distinct.

输出描述:

For each test case you should output the student records of which the grades are in the given interval [grade1, grade2] 
and are in non-increasing order.  Each student record occupies a line with the student‘s name and ID, separated by one space.  
If there is no student‘s grade in that interval, output "NONE" instead.

输入例子:

4
Tom CS000001 59
Joe Math990112 89
Mike CS991301 100
Mary EE990830 95
60 100

输出例子:

Mike CS991301
Mary EE990830
Joe Math990112

思路:简单模拟，n个学生信息包括学生姓名，科目代号，分数，给两个分数的左右区间，问分数在这两个区间的范围内的学生有哪些，有的话按照成绩的非递增顺序输出学生的姓名和科目。
没有输出“NONE”

#include<bits/stdc++.h>
using namespace std;

struct student {
    char name[11];
    int grade;
    char subject[11];
};
student res[101];
student stu[101];
bool cmp(student a, student b)
{
    return a.grade >= b.grade;
}
int main()
{
    //freopen("in.txt", "r", stdin);
    ios::sync_with_stdio(false), cin.tie(0);
    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
        cin >> stu[i].name >> stu[i].subject >> stu[i].grade;
    int left, right;
    cin >> left >> right;
    int cnt = 0;
    for (int i = 0; i < n; i++)
        if (stu[i].grade >= left && stu[i].grade <= right)
            res[cnt++] = stu[i];
    if (cnt == 0) {cout << "NONE"; return 0;}
    sort(res, res + cnt, cmp);
    for (int i = 0; i < cnt; i++)
        cout << res[i].name << " " << res[i].subject << endl;
    return 0;

}

View Code

List Grades (25)

标签：技术   syn   col   return   style   you   with   分享图片   amp   

原文地址：https://www.cnblogs.com/kuroko-ghh/p/10289921.html